Task: find a triangle with a smaller perimeter
I came across this task by accident. A friend of mine one year after graduating from the magistracy decided to study again and began to prepare for admission. So something just needs to be repeated and remembered, well, and sorted out with something new. Here she sat over some task, I passed by. The task seemed very simple (school level), but you need to think a little.
So, the task considered here sounds like this: given the angle and the point inside it. Through this point, draw segments having ends on the sides of the corner so that the resulting triangle has the smallest perimeter.
The problem is part of the proof of the Fagnano problem .
The first thoughts that come to mind are probably to build perpendiculars (as the shortest distance to the sides). Display point symmetrically with respect to and (we get points and )
Some may immediately be tempted to connect the intersection points of the perpendiculars and the sides of the angle. After which there is a false impression “I did it”, and it seems that- this is the same triangle.
All wrong. The fact that the two sides of the triangle are the shortest (perpendicular to the straight line) does not make the perimeter of the triangle minimal.
In fact, the search for the triangle with the smallest perimeter uses the statement: the shortest distance between two points is a straight line. Additional constructions should lead to the fact that all the lengths of the sides of the desired triangle are on a straight line. Connect the dots and . Line intersection points with the sides of the angle are the remaining desired vertices of the triangle.
and are medians and heights (dot symmetrically displayed relative to the sides of the corner and respectively, it means triangles and - isosceles. It is seen that the perimeter of the triangle equal to the length of the segment . A triangle with a smaller perimeter is found.
Let's take some other points ( and ) on the sides of the corner.
Perimeter of this triangle turns out to be greater than the length of the segment .
That's all. Good luck to everyone!
So, the task considered here sounds like this: given the angle and the point inside it. Through this point, draw segments having ends on the sides of the corner so that the resulting triangle has the smallest perimeter.
The problem is part of the proof of the Fagnano problem .
The task of Fagnano itself is as follows:
All kinds of triangles are considered. tops , and which lie on the sides , and acute triangle respectively. Prove that of all triangles DEF, the orthocentric triangle of the triangle has the smallest perimeter.
Orthocentric triangle
An orthocentric triangle (orthotriangle) is a triangle whose vertices serve as the base of the heights of the original triangle.
The first thoughts that come to mind are probably to build perpendiculars (as the shortest distance to the sides). Display point symmetrically with respect to and (we get points and )
Some may immediately be tempted to connect the intersection points of the perpendiculars and the sides of the angle. After which there is a false impression “I did it”, and it seems that- this is the same triangle.
All wrong. The fact that the two sides of the triangle are the shortest (perpendicular to the straight line) does not make the perimeter of the triangle minimal.
In fact, the search for the triangle with the smallest perimeter uses the statement: the shortest distance between two points is a straight line. Additional constructions should lead to the fact that all the lengths of the sides of the desired triangle are on a straight line. Connect the dots and . Line intersection points with the sides of the angle are the remaining desired vertices of the triangle.
and are medians and heights (dot symmetrically displayed relative to the sides of the corner and respectively, it means triangles and - isosceles. It is seen that the perimeter of the triangle equal to the length of the segment . A triangle with a smaller perimeter is found.
Let's take some other points ( and ) on the sides of the corner.
Perimeter of this triangle turns out to be greater than the length of the segment .
That's all. Good luck to everyone!